# How to perform a Gaussian functional integral?

I'm completely beginner to the quantum field theory and try to learn the basics of functional integrals. However, I could not understand clearly. Could someone please explain the idea with the help of the following example: $$$$\mathcal{Z} = \int[dE][dx] \exp \left[ - \int_{0}^{\beta} d\tau \left( \frac{1}{2}m\dot{x}^{2} + \frac{1}{2}m\omega_{0}^{2}x^{2} - eEx + \frac{1}{2g}(\dot{E}^{2} + \omega_{LC}^{2}E^{2}) \right) \right]$$$$ Let's say I want to perform the integral over $$[dx]$$ of partition function $$\mathcal{Z}$$. Could someone explain with some initial steps how to do that?

• Apr 11, 2022 at 7:31
• @Qmechanic Sorry! I'm not sure how to link my question to the answer you suggested Apr 11, 2022 at 7:42
• a few people have found the DETAILS section in physics.stackexchange.com/a/384919/83405 quite useful, it goes through the reasoning for evaluating quantum-mechanical path integrals quite carefully and slowly. Apr 11, 2022 at 18:24

In general for a real symmetric N x N matrix A and vector x we have

$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \cdots \int_{-\infty}^{+\infty} d x_{1} d x_{2} \cdots d x_{N} e^{-\frac{1}{2} x \cdot A \cdot x+J \cdot x}=\left(\frac{(2 \pi)^{N}}{\operatorname{det}[A]}\right)^{\frac{1}{2}} e^{\frac{1}{2} J \cdot A^{-1} \cdot J}$$

Which you can derive by diagonalizing A with an orthogonal transformation i.e $$A=O^{-1} \cdot D \cdot O$$

Set $$y_{i}=O_{i j} x_{j}$$ and the expression in the exponential becomes $$-\frac{1}{2} y \cdot D \cdot y+(O J) \cdot y$$

The integrals over $$x_i$$ just become integrals over $$y_i$$ and so we just have integrals of the form $$\int_{-\infty}^{+\infty} d y_{i} e^{-\frac{1}{2} D_{i i} y_{i}^{2}+(O J)_{i} y_{i}}$$

and plugging into the general result for these types of integrals, namely $$\int_{-\infty}^{+\infty} d x e^{-\frac{1}{2} a x^{2}+J x}=\left(\frac{2 \pi}{a}\right)^{\frac{1}{2}} e^{J^{2} / 2 a}$$

and using the fact that $$(O J) \cdot D^{-1} \cdot(O J)=J \cdot O^{-1} D^{-1} O \cdot J=J \cdot A^{-1} \cdot J$$

we get the right hand side of the first equation $$\left(\frac{(2 \pi)^{N}}{\operatorname{det}[A]}\right)^{\frac{1}{2}} e^{\frac{1}{2} J \cdot A^{-1} \cdot J}$$

Source: Zee QFT

Let's start with $$\mathcal{Z}=\int \exp(-x\cdot D\cdot x-2x\cdot y) \mathcal{D}[x] ,$$ where $$\cdot$$ represents an integral contraction (see below), and $$D$$ is some operator/kernel. For the example above, it is something like $$D\propto \partial_t^2-\omega_0^2$$, after partial integration of the $$\dot{x}^2$$-term.

There are basically three steps: 1) complete the square: $$\mathcal{Z}=\int \exp(-(x+y\cdot D^{-1})\cdot D\cdot (x+D^{-1}\cdot y)+y\cdot D^{-1}\cdot y) \mathcal{D}[x] .$$ 2) shift $$y\cdot D^{-1}$$ into the field variable: $$\mathcal{Z}=\int \exp(-x'\cdot D\cdot x'+y\cdot D^{-1}\cdot y) \mathcal{D}[x'] .$$ 3) evaluate the Gaussian integral: $$\mathcal{Z}=\exp(y\cdot D^{-1}\cdot y)\int \exp(-x'\cdot D\cdot x') \mathcal{D}[x']=\exp(y\cdot D^{-1}\cdot y)\frac{1}{\sqrt{\det\{D\}}} .$$

Definition of integral contraction: $$x\cdot D\cdot x = \int x(t) D(t,t') x(t') dtdt' .$$

• Thanks! But I did not understand how you arrive at the top most equation? What happen to the integral over $d\tau$ Apr 11, 2022 at 12:50
• It is included in the $\cdot$ --- see edit. Apr 11, 2022 at 12:51
• I somehow get your abstract answer :) . Suppose now, from purely physical point of view if limit $\omega_{LC}<< \omega_{0}$ is taken. Would I get the again harmonic oscillator action? Apr 11, 2022 at 13:06
• Can't say. You'll have to work through it. Apr 12, 2022 at 2:33